The values of p and q work out as -2 and 4 respectively thus complying with the given conditions.
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
Their values work out as: a = -2 and b = 4
Points: (1, 2) and (9, 6) Midpoint: (5, 4) Slope: 1/2 Perpendicular slope: -2 Perpendicular bisector equation: y-4 = -2(x-5) => y = -2x+14 Therefore: k = -2 thus satisfying the given bisector equation
Form a simultaneous equation with chord and circle and by solving it:- Chord makes contact with circle at: (-1, 4) and (3, 8) Midpoint of chord: (1, 6) Slope of chord: 1 Slope of perpendicular bisector: -1 Perpendicular bisector equation: y-6 = -(x-1) => y = -x+7
Equation of line: y = x+5 Equation of circle: x^2 +4x +y^2 -18y +59 = 0 The line intersects the circle at: (-1, 4) and (3, 8) Midpoint of line (1, 6) Slope of line: 1 Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -1(x-1) => y = -x+7 Perpendicular bisector equation in its general form: x+y-7 = 0
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
Their values work out as: a = -2 and b = 4
Points: (1, 2) and (9, 6) Midpoint: (5, 4) Slope: 1/2 Perpendicular slope: -2 Perpendicular bisector equation: y-4 = -2(x-5) => y = -2x+14 Therefore: k = -2 thus satisfying the given bisector equation
Form a simultaneous equation with chord and circle and by solving it:- Chord makes contact with circle at: (-1, 4) and (3, 8) Midpoint of chord: (1, 6) Slope of chord: 1 Slope of perpendicular bisector: -1 Perpendicular bisector equation: y-6 = -(x-1) => y = -x+7
Equation of line: y = x+5 Equation of circle: x^2 +4x +y^2 -18y +59 = 0 The line intersects the circle at: (-1, 4) and (3, 8) Midpoint of line (1, 6) Slope of line: 1 Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -1(x-1) => y = -x+7 Perpendicular bisector equation in its general form: x+y-7 = 0
True. (Apex)
Chord equation: y = x+5 Circle equation: x^2 +4x +y^2 -18y +59 = 0 Chord end points: (-1, 4) and (3, 8) Chord midpoint: (1, 6) Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -1(x-1) => y = -x+7
Chord equation: y = x+5 Circle equation: x^2 +4x +y^2 -18y +59 = 0 Both equations intersect at: (-1, 4) and (3, 8) which are the endpoints of the chord Midpoint of the chord: (1, 6) Slope of chord: 1 Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -(x-1) => y = -x+7
The slope of the line is 1/4 So the values are t = -2 and v = 4 Because they satisfy the equation: (v-2)/6-t = 2/8 = 1/4
If: y = 5x +10 and y = x^2 +4 Then: x^2 +4 = 5x +10 Transposing terms: x^2 -5x -6 = 0 Factorizing the above: (x-6)(X+1) = 0 meaning x = 6 or x = -1 Therefore by substitution endpoints of the line are at: (6, 40) and (-1, 5) Midpoint of line: (2.5, 22.5) Slope of line: 5 Perpendicular slope: -1/5 Perpendicular bisector equation: y-22.5 = -1/5(x-2.25) => 5y = -x+114.75 Perpendicular bisector equation in its general form: x+5y-114.75 = 0
False. 1). The proposed equation y=mx suggests that the chord's right bisector has no y-intercept, i.e. passes through the origin. This is interesting, and appears plausible, and I'm willing to acknowledge that this aspect of it is true. But ... 2). If the slope of the chord is 'm', then the slope of its right bisector is not also 'm'. If it were, that would make the chord and its bisector parallel, which would be pretty silly. The slope of any line perpendicular to the chord, including its right bisector, has to be '-1/m'. The equation of the chord's right bisector is: Y = -X/m .
It would be perpendicular to a line with the equation Y = 1/8 X.